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Question
Which of the following functions from
to itself are bijections?
Options
\[f\left( x \right) = \frac{x}{2}\]
\[g\left( x \right) = \sin\left( \frac{\pi x}{2} \right)\]
\[h\left( x \right) = |x|\]
\[k\left( x \right) = x^2\]
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Solution
\[\left( a \right) \text{Range of f}=\left[ \frac{- 1}{2}, \frac{1}{2} \right]\neq A\]
So, f is not a bijection.
\[\left( b \right) \text{Range }=\left[ \sin\left( \frac{- \pi}{2} \right), \sin\left( \frac{\pi}{2} \right) \right]=\left[ - 1, 1 \right]=A\]
So, g is a bijection.
\[\left( c \right) h\left( - 1 \right) = \left| - 1 \right| = 1\]
\[\text{ and } h\left( 1 \right) = \left| 1 \right| = 1\]
\[\Rightarrow-1 \text {and 1 have the same images}\]
So, h is not a bijection.
\[\] \[\left( d \right) k\left( - 1 \right) = \left( - 1 \right)^2 = 1\]
\[\text{and } k \left( 1 \right) = \left( 1 \right)^2 = 1\]
\[\Rightarrow-1 \text{and 1 have the same images}\]
So, k is not a bijection.
So, the answer is (b)
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